The following is taken from: $\textit{Partil Difgferential Control Theory Vo 1: Mathematical tools}$ by J F. Pommaret $\color{Green}{Background:}$
$\textbf{Definition 1.49.}$ If $M$ is a module over a ring $A,$ a $\textit{system of generators}$ of $M$ over $A$ is a family $\{x_i\}_{i\in I}$ of elements of $M$ such that any element $x\in M$ can be written $x=\sum_{i\in I}a_ix_i$ with $a_i\in A$ and $a_i=0$ for almost all $i,$ that is for all $i$ with a finite number of exceptions. In that case, we shall use to write $M=(\{x_i\}_{i\in I}).$ In particular, $M$ is $\textit{finitely generated}$ if it has a finite system of generators and $\textit{monogenic}$ if it can be generated by a single element.
$\textbf{Definition 1.50.}$ $M$ is call a $\textit{free module}$ if it has a $\textit{basis},$ that is a system of generators linearly indeoendent over $A.$ When $M$ is free, the number of generators in a basis is called the $\textit{rank}$ of $M$ and does not depend on the basis if it is finite. Indeed, if $\{x_i\}$ and $\{y_j\}$ are two bases and $y_j=\sum {a^i}_jx_i,$ then $\text{det}({a^i}_j)$ must be invertible in $A.$ If $M$ is free of rank $n,$ then $M\cong A^n.$
$\textbf{Proposition 1.55.}$ If $A$ is a noetherian ring and $M$ is a finitely generated $A-$module, then $M$ is a noetherian module.
$\textit{Proof.}$ Applying the preceding lemma (Lemma 1.54) to the short exact sequence $0\to A^{r-1}\to A^r\to A\to 0$ where the right epimorphism is the projection onto one factor, we deduce by induction that $A''$ is a noetherian module. Now, if $M$ is generated by $\{x_1,\ldots,x_n\}$ then there is an epimorphism $A^n\to M:(1,0,\ldots,0)\to x_1,\ldots,(0,\ldots,0,1)\to x_n.$ According to the preceding lemma, $M$ is a noetherian module too.
$\textbf{Remark 1.56.}$ In the situation of the preceding proposition, we notice that the kernel of the projection $A^n\to M$ is also a noetherian module according to the preceding lemma and thus finitely generated too. Accordingly, we may find an exact sequence of the type:
$$A^m\to A^n\to M\to 0.$$
In this case we say that $M$ is $\textit{finitely presented}$ or $\textit{presentable}$ if we do not specify the presentation. In general, a finitely generated module $M$ is said to be $\textit{coherent}$ if the kernel of an arbitrary homomorphism $A^n\to M$ is also finitely generated.
$\color{Red}{Questions:}$
In Remark 1.56 above in regards to the projection map $A^n\to M,$ we have a module $A$ of rank $n,$ meaning it has $n$ number of basis elements and is being projected to the module $M.$ So the rank of $M$ would be less than or equal to $M.$ But in the case of the short exact sequence: $A^m\to A^n\to M\to 0,$ the projection map only refer to the component $A^n\to M$ irrespective if in the component map $A^m\to A^n,$ $m\neq m?$
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